Explore Measurement Concepts with Our Examples and Questions

Consider the model
${𝑋}_{𝑖}\ast$
In practice we measure ${𝑋}_{𝑖}\ast$ by ${𝑋}_{𝑖}$ such that
a) ${𝑋}_{𝑖}\ast <!--\; \ast \; -->+3$
b) $5{𝑋}_{𝑖}\ast <!--\; \ast \; -->$
What will be the effect of these measurement errors on estimates of true ${𝛽}_0$ and ${𝛽}_1$? In terms of biasedness, BLUE, consistency, efficiency?

If $F$ is a continuous distribution function on $(\mathbb{R},\mathcal{B},{\mathrm{\mu <!--\; \mu \; -->}}_{\mathcal{L}})$ with distribution ${\mathrm{\mu <!--\; \mu \; -->}}_F$, use Fubini's theorem to show that
1. ${\int <!--\; \int \; -->}_{\mathbb{R}}F(x)d{\mathrm{\mu <!--\; \mu \; -->}}_F(x)=\frac{1}{2}$
2. If $X_1,X_2$ are i.i.d random variables with common distribution $F$, then $P(\{X_1\le <!--\; \le \; -->X_2\})=1/2$ and $\text{E}(F(X_1))=1/2$.

My Attempt:
I don't really understand bs_math's answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.
For example, I don't understand what's going on in line 4 of bs_math's answer.

Let $f_j\in <!--\; \in \; -->L^P\cap <!--\; \cap \; -->L_{loc}^1$ and $g\in <!--\; \in \; -->L^q$ where $p,q$ are conjugates and $||f_j|{|}_p<<!--\; <\; -->M$ for all $j$ . I have that ${\int <!--\; \int \; -->}_S{f}_j\to <!--\; \to \; -->0$ for all $S\subset <!--\; \subset \; -->A$ (measurable) where $A$ is a compact subset of $\mathbb{R}$ and want to show ${\int <!--\; \int \; -->}_A{f}_{j}g\to <!--\; \to \; -->0$.
Since $A$ is bounded, $g\in <!--\; \in \; -->L^q⟹<!--\; ⟹\; -->g\in <!--\; \in \; -->L^{\mathrm{\infty <!--\; \infty \; -->}}$ so I want to write down something like $|{\int <!--\; \int \; -->}_A{f}_{j}g|\le <!--\; \le \; -->||g|{|}_{L^{\mathrm{\infty <!--\; \infty \; -->}}}|\int <!--\; \int \; -->f_j|$. But this is false. I can only use the tools that I know if I control $\int <!--\; \int \; -->|f|$, but I've already come up with counterexamples to show we have no control over that so considering $\int <!--\; \int \; -->|f_{j}g|$ is useless. Another idea was to use Cauchy schwarz but we need not have $\int <!--\; \int \; -->f_j^2\to <!--\; \to \; -->0$

If $f=f(x)$ is absolutely continuous on [0,1] such that $f^{\prime }\in <!--\; \in \; -->L^2([0,1])$ and $f(0)=0$. Prove then that
$\underset{x\to <!--\; \to \; -->0^{+}}{lim}\frac{f(x)}{x^{\frac{1}{2}}}=0$
The hint says to use Fundamental theorem of calculus but I don't see where to use that.
I know if $f^{\prime }\in <!--\; \in \; -->L^2([0,1])$ then
$({\int <!--\; \int \; -->}_{[0,1]}|f^{\prime }(x){|}^{2}dx{)}^{\frac{1}{2}}<+\mathrm{\infty <!--\; \infty \; -->}.$
I don't know what to do from here though. Any hints would be appreciated. Am I merely using L'Hôpital's rule, since I get 0/0?

I will appreciate any help.
Consider the following function:
$g\left(x\right)=\{\begin{array}{ll}\frac{1}{x},& \text{if}x\in <!--\; \in \; -->(1,\mathrm{\infty <!--\; \infty \; -->})\\ 0,& \text{if}x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1]\end{array}$
Determine the set $\{p\in <!--\; \in \; -->[0,\mathrm{\infty <!--\; \infty \; -->}]:g\in <!--\; \in \; -->{\mathcal{L}}^p\left(\mathrm{\lambda <!--\; \lambda \; -->}\right)\}$. I know that the set is equal to $(1,\mathrm{\infty <!--\; \infty \; -->}]$. But I am not sure how to calculate the cases $p=0$ and $p=\mathrm{\infty <!--\; \infty \; -->}$.
In my textbook we have the following definitions for $p=0$ and $p=\mathrm{\infty <!--\; \infty \; -->}$:
${\mathcal{L}}^{\mathrm{\infty <!--\; \infty \; -->}}\left(\mathrm{\mu <!--\; \mu \; -->}\right)=\{f\in <!--\; \in \; -->\mathcal{M}\left(\mathcal{E}\right):\mathrm{\exists <!--\; \exists \; -->}R>0:|f|\le <!--\; \le \; -->R\mathrm{\mu <!--\; \mu \; -->}\text{-almost everywhere}\}$
and
${\mathcal{L}}^0\left(\mathrm{\mu <!--\; \mu \; -->}\right)=\{f\in <!--\; \in \; -->\mathcal{M}\left(\mathcal{E}\right):\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\mu <!--\; \mu \; -->}(\{|f|\ge <!--\; \ge \; -->t\})=0\}$
where $\mathcal{M}\left(\mathcal{E}\right)=\{f:X\to <!--\; \to \; -->\mathbb{R}:f\text{is}\mathcal{E}\text{-}\mathcal{B}\left(\mathbb{R}\right)\text{- measurable}\}$.

Let $A\subset <!--\; \subset \; -->\mathbb{R}$ be Lebesgue measurable such that $0<\mathrm{\lambda <!--\; \lambda \; -->}(A)<\mathrm{\infty <!--\; \infty \; -->}$ and consider the function $f:\mathbb{R}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$, $f(x)=\mathrm{\lambda <!--\; \lambda \; -->}(A\cap <!--\; \cap \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x))$. I would like to compute $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)$.
I have already proved that f is continuous (I have in fact showed that it is Lipschitz continuous), but I am not really sure if this allows me to say that $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)=\mathrm{\lambda <!--\; \lambda \; -->}(A\cap <!--\; \cap \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->}))=\mathrm{\lambda <!--\; \lambda \; -->}(A)$. Intuitively, I am sure that this is right, but I am not sure whether the fact that $f$ is continuous is enough to write this. I feel that I am somehow assuming that the Lebesgue measure $\mathrm{\lambda <!--\; \lambda \; -->}$ is continuous in some sense. Could you please tell me if my conclusion is right and if the mere continuity of $f$ is enough for it?

I have a prediction $f(x)$ of some continuous process variable, based on an input variable $x$ (think: location). The prediction is incorrect, with the error being normal distributed with expected value $\mathrm{\mu <!--\; \mu \; -->}$ and standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}$.
Hence, the probability density function of $f(x)$ should be
$\frac{1}{\mathrm{\sigma <!--\; \sigma \; -->}\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{-<!--\; -\; -->\frac{((f(x)-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})/\mathrm{\sigma <!--\; \sigma \; -->}{)}^2}{2}}$
Is this correct? (No it is not, see answer below)
Now, I have a measurement $m$ of the process variable, based on an unknown input variable $x_m\cdot <!--\; \cdot \; -->m$ is assumed to be correct, but quantized to integral numbers.
Given a set of $x_i$ together with their predictions $f(x_i)$, how can I compute a probability that $x_m$ was in the vicinity of $x_i$?
I apologize if the question makes no sense.

I have a bunch of points where the best circle is fitted into them. The algorithm is based on the least squares approach to fit a circle.
My question is, how would error analysis be performed. I am thinking of something like:
$\underset{i}{\sum <!--\; \sum \; -->}(r_i-<!--\; -\; -->r_{(\text{actual})}{)}^2$ where $r_i$ are the distances from the center of the fitted circle to the point, and $r_{actual}$ is the actual radius of the circle.
Any suggestions or insights on how to proceed?

Let $E\subset <!--\; \subset \; -->{\mathbb{R}}^n$ and $O_i=\{x\in <!--\; \in \; -->{\mathbb{R}}^n\mid <!--\; \mid \; -->d(x,E)<\frac{1}{i}\},i\in <!--\; \in \; -->\mathbb{N}$. Show that if $E$ is compact, then $m(E)=\underset{i\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}m(O_i)$. Does the statement hold if $E$ is closed, but not bounded and if $E$ is open, but bounded?
What I got was that since $\underset{i}{\bigcap <!--\; \bigcap \; -->}{O}_i=E$ and $O_1\supset <!--\; \supset \; -->O_2\supset <!--\; \supset \; -->\dots <!--\; \dots \; -->$, then we have that
$m(E)=m(\underset{i}{\bigcap <!--\; \bigcap \; -->}{O}_i)=\underset{i\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}m(O_i)$
by the continuity of the Lebesgue measure. How is the fact needed that $E$ is compact here?

Let $\mathrm{\gamma <!--\; \gamma \; -->}:[0,1]\to <!--\; \to \; -->{\mathbb{S}}^2$ be a smooth curve. Consider the following set
$\underset{t\in <!--\; \in \; -->[0,1]}{\bigcup <!--\; \bigcup \; -->}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(\mathrm{\gamma <!--\; \gamma \; -->}(t))\subset <!--\; \subset \; -->{\mathbb{R}}^3$
My question is: does this set have Hausdorff dimension at most 2, and as a result it has zero 3-dimensional Lebesgue measure?. Note that if the smoothness condition is removed then the answer is trivially false.

There is a question in a book that I am trying to solve.
"A man usually rides his bike 1 kilometers per hour, yet the wind slows him to 6.76 kilometers for 26 minutes and 5.55 kilometers for 10. How long until he gets home 11.54 kilometers away?"

Assume $f_1,\dots <!--\; \dots \; -->,f_N:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$. If $\int <!--\; \int \; -->|f_n|\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}$ for all $n=1,\dots <!--\; \dots \; -->,N$, then we always have
$\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}\int <!--\; \int \; -->f_n\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{f}_n\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}.$
Now we assume $f_1,\dots <!--\; \dots \; -->,f_N:\mathbb{R}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->}]$ with $\int <!--\; \int \; -->f_1\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}=\mathrm{\infty <!--\; \infty \; -->}$. I would like to ask if
$\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}\int <!--\; \int \; -->f_n\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{f}_n\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}$
still holds. Thank you so much for your explanation!

Let $E\subset <!--\; \subset \; -->\mathbb{R}$ be a Borel set and suppose that $(f_k{)}_{k\in <!--\; \in \; -->\mathbb{N}}$ is a sequence of Lebesgue measurable functions defined on $E$. Assume that $\underset{k\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}{f}_k(x)=f(x)$ for almost every $x$ and that there exists a Lebesgue integrable function $g$ such that $|f_k(x)|\le <!--\; \le \; -->g(x)$ for almost every $x\in <!--\; \in \; -->E$ and every $k\in <!--\; \in \; -->\mathbb{N}$. Fix $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ and define $A_k^{\mathrm{\epsilon <!--\; \epsilon \; -->}}=\{x\in <!--\; \in \; -->E:|f_k(x)-<!--\; -\; -->f(x)|\ge <!--\; \ge \; -->\mathrm{\epsilon <!--\; \epsilon \; -->}\}$. Is it true that
$\underset{j\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}\left|\underset{k\ge <!--\; \ge \; -->j}{\bigcup <!--\; \bigcup \; -->}{A}_k^{\mathrm{\epsilon <!--\; \epsilon \; -->}}\right|=0?$
($|\cdot <!--\; \cdot \; -->|$ is the Lebesgue measure)
The only thing that I noticed is that $|A_k^{\mathrm{\epsilon <!--\; \epsilon \; -->}}|\underset{k\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{\overset{}{\to }}0$, as a consequence of Lebesgue's dominated convergence theorem.
Does anyone know if this is true or not (and why?)?

Let's say we conduct a random experiment. The possible outcomes may be too complicated to describe, so we instead take some measurement (in terms of real numbers) from it which are of interest to us. Mathematically we have defined a random variable X from the sample space/measurable space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A})$ to $(\mathbb{R},\mathcal{B})$ where $\mathcal{B}$ is the Borel sigma field. We then choose an appropriate probability for each Borel set, via a distribution function. Then, how does that distribution function determine the probability space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},P)$ in theory or proves its existence? What result is implicitly used here?
I understand the probability space on $(\mathbb{R},\mathcal{B})$ is well defined, and that is what practically concerns us, but in theory we also have a probability space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},P)$ which pushes forward its measure to the Borel sets. Hence given the distribution on $X$ we are pulling back to $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A})$ via $P(X\in <!--\; \in \; -->B)=P_X(B)$. But what theorem guarantees that such a space exists?
Edit for clarification:For example, let us take weather $\mathrm{\Omega <!--\; \Omega \; -->}$ of a city as the original sample space and the recorded temperature $X$ as the random variable. Suppose the distribution of $X$ is decided upon, based on empirical data. That's fine and now we can answer questions like 'what is $P(a<X<b)$'. We may not care about the probability space on $\mathrm{\Omega <!--\; \Omega \; -->}$ since all the probabilities we wish to know relate to $X$, but how do we know for sure that there exists a probability space on $\mathrm{\Omega <!--\; \Omega \; -->}$ in the first place, and further, how do we know that a probability space on $\mathrm{\Omega <!--\; \Omega \; -->}$ exists which would push its probability to yield the distribution of $X$? Unless we know that there exists such a probability space in theory, we will not be able to talk about expectation of $X$, so the knowledge that such a space exists is crucial.
*Further edit: Note that we may not have an explicit complete mathematical model of the weather ever, but that does not bother us. We do have the ability however, of taking measurements of different types and thereby getting distributions related to temperature, pressure, wind speed, historical records etc. So practically we can answer questions related to these measurements. This is so what happens in practice I think. What I want is some mathematical theorem which assures me that there exists a model of the weather space, even though it is hard/difficult/impossible to find, which pushes its probability on to these distributions.

We know that a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra is either finite or not countable. Can I find an algebra (field of sets) that is countably infinite? I know that the proof for said theorem uses the fact that a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra is closed so countable unions, but I cant find an example.

Suppose $X$,$Y$ are real-valued random vectors. When can we say $E[g(X,Y)|X]$ is an a.e. continuous function in $X$? I have mostly convinced myself of this argument, save for one step where I want to argue $E[\frac{1}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i{X}_i^{\prime }{e}_i^2|(X_1,...,X_n{)}^{\prime }]\stackrel{p}{\to <!--\; \to \; -->}E[E\left[X_i{X}_i^{\prime }{e}_i^2\right]|(X_1,...,X_n{)}^{\prime }]=E\left[X_i{X}_i^{\prime }{e}_i^2\right]$ by appealing to the weak law of large numbers and the continuous mapping theorem.

if I know that a function $f$$\in <!--\; \in \; -->L^2({\mathbb{R}}^d)$ is supported in a compact ball in ${\mathbb{R}}^d$, how do I prove it is in $L^1({\mathbb{R}}^d)$

From the square integrability I know that
for some constant $C$, then I know that $f(supp(f))$ is bounded, I need to show that it is closed as well to conclude that it is compact, but that will not be true in general if $f$ is not continuous.

How am I supposed to start given what I have?

I want to prove the following two parts of the question.
a) Show that $\{{\mathrm{\mu <!--\; \mu \; -->}}_n\}$ is tight if and only if for each $\mathrm{ϵ<!--\; ϵ\; -->}$ there is a compact set K such that ${\mathrm{\mu <!--\; \mu \; -->}}_n(K)>1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$ for all n.
b) Show that $\{{\mathrm{\mu <!--\; \mu \; -->}}_n\}$ is tight if and only if each of the k sequences of marginal distributions is tight on the line.
I know that
A sequence $\{{\mathrm{\mu <!--\; \mu \; -->}}_n\}$ of probability measures is tight if for every ϵ there is a bounded rectangle A such that ${\mathrm{\mu <!--\; \mu \; -->}}_n(A)>1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$ for all n.
By this definition, we know that a bounded rectangle A in $R^k$ is compact. This implies $\{{\mathrm{\mu <!--\; \mu \; -->}}_n\}$ is tight.
This does not looks good to me.The question is basically asking to prove the definition.
Can anyone help to solve this question?

Let $\mathrm{\mu <!--\; \mu \; -->}$ be a complex measure on ${\mathbb{R}}^n$ and $f\in <!--\; \in \; -->L^1(\mathrm{\mu <!--\; \mu \; -->})$ such that $f(x)\ge <!--\; \ge \; -->c>0$ ( for some constant $c>0$ ) a.e. $x\in <!--\; \in \; -->{\mathbb{R}}^n$. Then is it true that
$|{\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->}|\ge <!--\; \ge \; -->|{\int <!--\; \int \; -->}_{{\mathbb{R}}^n}cd\mathrm{\mu <!--\; \mu \; -->}|=c|\mathrm{\mu <!--\; \mu \; -->}({\mathbb{R}}^n)|\text{ ? }$?

My first thinking process is to break each integral first in its real and imaginary part and then their corresponding positive and negative parts, i.e.
${\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->}=Re({\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->})+iIm({\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->})$
$=R{e}^{+}({\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->})-<!--\; -\; -->R{e}^{-<!--\; -\; -->}({\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->})+iI{m}^{+}({\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->})-<!--\; -\; -->iI{m}^{-<!--\; -\; -->}({\int <!--\; \int \; -->}_{{\mathbb{R}}^n}fd\mathrm{\mu <!--\; \mu \; -->})$
and then wanted to look at corresponding decompositions of the measure and estimate on each such segements. Also note that a complex measure by definition gives $|\mathrm{\mu <!--\; \mu \; -->}({\mathbb{R}}^n)|<\mathrm{\infty <!--\; \infty \; -->}$ . Now how to proceed from here.
Can someone please help?

Given a probability space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},P)$ and a random variable $X$ defined on it, s.t. $X\ge <!--\; \ge \; -->0$, I want to show that the mapping
$(\mathrm{\omega <!--\; \omega \; -->},t)\mapsto <!--\; \mapsto \; -->{\mathbf{1}}_{\{X(\mathrm{\omega <!--\; \omega \; -->})\ge <!--\; \ge \; -->t\}}$
is measurable on the measure space $(\mathrm{\Omega <!--\; \Omega \; -->}\times <!--\; \times \; -->[0,\mathrm{\infty <!--\; \infty \; -->}),\mathcal{F}\times <!--\; \times \; -->\mathcal{B}([0,\mathrm{\infty <!--\; \infty \; -->})),P\otimes <!--\; \otimes \; -->\mathrm{\lambda <!--\; \lambda \; -->})$, for $\mathrm{\lambda <!--\; \lambda \; -->}$ denoting the Lebesgue measure.

My thoughts were to simply use the definition and look at the preimages of this mapping. One would then have, e.g.
${\mathbf{1}}_{\{X(\mathrm{\omega <!--\; \omega \; -->})\ge <!--\; \ge \; -->t\}}^{-<!--\; -\; -->1}(\{0\})=\{(\mathrm{\omega <!--\; \omega \; -->},t):X(\mathrm{\omega <!--\; \omega \; -->})<t\}=\{(\mathrm{\omega <!--\; \omega \; -->},t):\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->X^{-<!--\; -\; -->1}([0,t))\}.$
Since $X$ is a random variable, it is measurable, so $X^{-<!--\; -\; -->1}([0,t))\in <!--\; \in \; -->\mathcal{F}$. However, I cannot see at the moment why $\{(\mathrm{\omega <!--\; \omega \; -->},t):\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->X^{-<!--\; -\; -->1}([0,t))\}$ belongs to the product $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra, since one coordinate depends on the other. I'm grateful for any hints.